Question

Question #f639b

Answer 1

Use the Ratio Test (it's usually a good idea whenever factorials are involved) to show that this series converges.

Explanation:

Let `a_{n}=(2/n)^{n}*n!`.

We want to know whether the series `sum_{n=1}^{infty}a_{n}` converges or not.

To apply the Ratio Test, we consider the sequence `|a_{n+1}/a_{n}|`. In this example, all the terms in the original sum are positive, so we may dispense with the absolute value signs and note that `a_{n+1}/a_{n}=(2/(n+1))^(n+1) * (n+1)!*(n/2)^(n)*1/(n!)`.

This expression simplifies to:

`a_{n+1}/a_{n}=2(n/(n+1))^(n)`.

It is well known that `((n+1)/n)^{n}=(1+1/n)^{n}->e` as `n-> infty`. Therefore, `a_{n+1}/a_{n} -> 2/e approx 0.73576<1` as `n-> infty`.

By the Ratio Test, this is enough to imply that the series `sum_{n=1}^{infty}(2/n)^{n}*n!` converges.

Answer 2

See below.

Explanation:

Using the Stirling asymptotic formula

`n! approx sqrt(2pin)(n/e)^n` we have

`sum_(n=1)^oo(2/n)^2n! approx sum_(n=1)^oo(2/n)^2sqrt(2pin)(n/e)^n = sum_(n=1)^oo sqrt(2pin)(2/e)^n` and

`sum_(n=1)^oo sqrt(2pin)(2/e)^n le sum_(n=1)^oo (n+1)(2/e)^n`

and from

`sum (k+1)x^k = d/dx(sum x^(k+1))` the convergence for `sum (k+1)x^k` is assured if `abs x < 1` or in our case

`2/e < 1` and as a consequence

`sum_(n=1)^oo(2/n)^2n!` converges.